is this correct $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$? I think i might have seen this result somewhere $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$ but cant remember for sure.
Is there a name for sums having limits in both lower and upper bound?
is there anything similar for other numbers e.g. $\lim_{ n \to \infty} \sum_{k=3^n}^{3^{n+1}} \frac{1}{k}= \ln 3 ?$
 A: Yes, the result is correct, because it is a Riemann sum.  Recalling that $$\int_{x=a}^b f(x) \, dx = \lim_{n \to \infty} \sum_{k=0}^{n-1} f\left(a + \frac{b-a}{n} k\right) \frac{b-a}{n},$$ with the choice $a = 1$, $b = 2$, $n = 2^m$, $f(x) = \frac{1}{x}$, we obtain $$\int_{x=1}^2 \frac{1}{x} \, dx = \lim_{m \to \infty} \frac{1}{2^m} \sum_{k=0}^{2^m - 1} \frac{1}{1 + k/2^m} = \lim_{m \to \infty} \left(-\frac{1}{2^{m+1}} + \sum_{k=0}^{2^m} \frac{1}{2^m + k}\right)  = \lim_{m \to \infty} \sum_{k=2^m}^{2^{m+1}} \frac{1}{k}.$$
I leave it to you to see if you can extend this method to discover forms that evaluate to $\log 3$, $\log 4$, etc.
A: For $a>1$ and $n\in\Bbb N$ we have
$$\int_{a^n}^{a^{n+1}+1}\frac{dx}x<\sum_{k=a^n}^{a^{n+1}}\frac 1k<\int_{a^n}^{a^{n+1}+1}\left(\frac1x+\frac1{a^n}\right)dx$$
A: Let $\alpha=\lim\limits_{n\to\infty}\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$
The following is taken from this answer:
In this answer, It is shown that $\left(1+\frac1n\right)^n$ increases and $\left(1+\frac1n\right)^{n+1}$ decreases to $e$. Thus,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}\le\dots\le\left(1+\frac1{n+k}\right)^{n+k}\tag{5}
$$
and
$$
\left(1+\frac1n\right)^{n+1}\ge\left(1+\frac1{n+1}\right)^{n+2}\ge\dots\ge\left(1+\frac1{n+k}\right)^{n+k+1}\tag{6}
$$
Putting $(5)$ and $(6)$ together yields, for $0\le k\le n$,
$$
\left(1+\frac1{n+k}\right)^{\frac{n}{n+1}}\le\left(1+\frac1{n+k}\right)^{\frac{n}{n+k}\frac{n+k+1}{n+1}}\le\left(1+\frac1n\right)^\frac{n}{n+k}\le\left(1+\frac1{n+k}\right)\tag{7}
$$
Therefore,
$$
\begin{align}
e^\alpha
&=\lim_{n\to\infty}\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{8}
\end{align}
$$
and by $(7)$
$$
\begin{align}
\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}
&\le\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\dots\left(1+\frac1{2n}\right)\\
&=\frac{2n+1}{n+1}\tag{9}
\end{align}
$$
Using the other direction of $(7)$, we get
$$
\left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1}
\le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)}\tag{10}
$$
By the Squeeze Theorem, $(8)$, $(9)$, and $(10)$ say
$$
e^\alpha=2\tag{11}
$$
which, by definition, means
$$
\alpha=\log(2)\tag{12}
$$
A: Let $S(A)=\sum _{n=1}^{n=A}\frac {1}{n}.\;$  We have $\frac {1}{n+1}=\int_n^{n+1}\frac {1}{n+1}dx<\int_n^{n+1}\frac {1}{x}dx<\int_n^{n+1}\frac {1}{n}dx=\frac {1}{n}.$ $$\text {Therefore }\quad -1+S(A)+\frac {1}{m+1}<\int_1^A\frac {1}{x}dx=\log A<S(A).$$ $$\text {Equivalently }\quad \log A<s(A)<(\log A)+1-\frac {1}{A+1}.$$ $$\text { Observe that }\quad \sum_{k=2^m}^{2^{m+1}}\frac {1}{k}=S(2^{m+1})-S(2^m-1).$$ Note : $S(A)-\log A$ converges to about $ 0.577...$ as $A\to \infty.$ This is known as Euler's constant.
